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maks197457 [2]
2 years ago
7

A 1.4-µC point charge is placed between the plates of a parallel plate capacitor. The charge experiences a force of 0.38 N. What

is the magnitude σ of the charge density on either plate of the capacitor?
Physics
1 answer:
Art [367]2 years ago
4 0

Answer:

2.4 * 10^-6 C/m²

Explanation:

The Force experimented by a particle in an electric field will be equal to:

F = E*q

Where E is the magnitude of the electric field and q is the charge of the particle.

The Electric field caused by a parallel plate capacitor will be:

E = σ/ε0

Where σ is the charge density of the plates of the capacitor and ε0 is the Vacuum permittivity constant.

We replace this expression in the expression for force and issolate σ and get that:

σ = F*ε0/q = 0.38 N * 8.854 * 10^-12 F/m / (1.4*10^-6C) = 2.4 * 10^-6 C/m²

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a steep-sided valley formed by the downward displacement of a block of the earth's surface between nearly parallel faults or fault systems.

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One kg of air is contained in a piston-cylinder system and it undergoes a Carnot cycle having an efficiency of 60%.The heat tran
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Answer is in the file below

tinyurl.com/wtjfavyw

4 0
2 years ago
A system of releases 125kJ of heat while 104kJ of work is done in the system. Calcilate the change om imternal energy (in kJ)
artcher [175]

Answer:

DU = 21 KJ

Explanation:

Given the following data;

Quantity of heat = 125 KJ

Work = 104 KJ

To find the change in internal energy;

Mathematically, the change in internal energy of a system is given by the formula;

DU = Q - W

Where;

DU is the change in internal energy.

Q is the quantity of energy.

W is the work done.

Substituting into the formula, we have;

DU = 125 - 104

DU = 21 KJ

4 0
2 years ago
Helicopters rotor blades, could spin at high speed of 510 rpm. Find the angular displacement in radian for 3 hour(s) operation
malfutka [58]

Answer:

The angular displacement of the blade is 576,871.2 radians

Explanation:

Given;

angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)

time of motion, t = 3 hours

The angular speed of the Helicopters rotor blades in radian per second is given as;

\omega = \frac{510 \ rev}{mins} *\frac{2 \pi \ rad}{1 \ rev} *\frac{1 \ min}{60 \ s}\\\\\omega =  53.414 \ rad/s

The angular displacement in radian is given as;

θ = ωt

where;

t is time in seconds

θ = (53.414)(3 x 60 x 60)\\

θ = 576,871.2 radians

Therefore, the angular displacement of the blade is 576,871.2 radians

5 0
2 years ago
By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what
Vinil7 [7]

Explanation:

1. Mass of the proton, m_p=1.67\times 10^{-27}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}

V=\dfrac{h^2}{2q_pm_p\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}

V = 45.83 volts

2. Mass of the electron, m_p=9.1\times 10^{-31}\ kg

Wavelength, \lambda_p=4.23\times 10^{-12}\ m

We need to find the potential difference. The relationship between potential difference and wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}

V=\dfrac{h^2}{2q_em_e\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}

V=6.92\times 10^{34}\ V

V = 84109.27 volt

Hence, this is the required solution.

7 0
3 years ago
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